Determine if the function is continuous to its domain

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

Yes, it is continuous to its domain

<h2>Explanation:</h2><h2></h2>

In order to find the domain of the function we need to get the restrictions:

1. From natural log:

$x>0$

2. From quotient:

$x\neq 0$

Matching these two restrictions the domain is:

$x\in (0,\infty)$

So the function is continuous to its domain because is defined for every x-value in the interval $(0, \infty)$ )

<h2>Learn more:</h2>

Functions: brainly.com/question/12891789

#LearnWithBrainly

You might be interested in

Four doughnuts and 4 coffees cost $12. The cost of the 4 coffees is half the cost of the 4 doughnuts. What is the cost of each c

tangare [24]

The cost of each coffee is $1.05

8 0

3 years ago

Show that if S1 and S2 are subsets of a vector space V such that S1 c S2 then span(S1) c span(S2). In particular, if S1 c S2 the

klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

If, additionally $S_2\subseteq S_1$ then reversing the roles of S1 and S2 in the previous proof, $span(S_2)\subseteq span(S_1)$ . Then $span(S_1)\subseteq span(S_2)\subseteq span(S_1)$ , therefore $span(S_1)=span(S_2)$ .